Geodesics in Curved Spacetime: What is the Significance of the ct Increment?

[Bob R]

[Moderator's note: this post has been spun off into its own thread.]

I'm a retired engineer trying to get my head around GR, its effects in our everyday non-relativistic world, and its reduction to Newtonian gravity. I hope this is not too much of a digression from the current string. As I understand it, the presence of mass/energy causes curvature in space-time and that curvature causes mass/energy to travel along a force-free geodesic. If I toss a ball five meters high and five meters horizontally it takes about 2 seconds to land. On a worldline the ct axis increment is (2 sec)*(3 e8 m/s) or about 6 e8 m. The xy plane space increment on the worldline is 5 m in the x and y directions. Is the ball traveling force free along the worldline? Is there any physical significance to the ct increment of 6 trillion meters or is this just a mathematical byproduct to be ignored? When one considers the Earth's movement in one day about the sun, is the Earth traveling in space along a geodesic arc of about (360 deg/365.24 days)? Is the ct component of this worldline (3 e8 m/sec)*(86,400 sec/day) and what does that mean? If a test particle is released at the Earth's position withe zero tangential velocity it will travel toward the sun. Is this also a geodesic path and does the geodesic path depend on the velocity of the test particle relative to the sun? Thanks for any understanding you can provide. Bob R.

 

[PeterDonis]

Is the ball traveling force free along the worldline?

Yes.

Is there any physical significance to the ct increment of 6 trillion meters

Yes; it gives you information about the spacetime geometry in the vicinity of the Earth. Here's a quick heuristic way of seeing how: we have two curves in spacetime: the worldline of the ball, and the worldline of someone on the surface of the Earth who moves from the ball's starting point to its ending point. (To make the scenario even simpler, we could assume that the ball moves purely vertically, so the person on the surface is just standing still.) The ball's worldline is straight (i.e., a geodesic); the worldline of the person on the surface of the Earth is not. Over a distance in spacetime of 6 trillion meters, the two curves deviate by a maximum of 5 meters. That's a very small relative deviation, and it indicates that, in the vicinity of the Earth's surface, the curvature of spacetime is very small.

When one considers the Earth's movement in one day about the sun, is the Earth traveling in space along a geodesic arc of about (360 deg/365.24 days)?

This is the arc in space, but not in spacetime. See below.

Is the ct component of this worldline (3 e8 m/sec)*(86,400 sec/day)

Yes, it is, and this means that in spacetime, the arc is a small segment of a helix, which goes around a "horizontal" (spatial) arc of $(360/365.24) 1.5 \times 10^{11} \approx 10^{11}$ meters in a "vertical" (time) distance of $(86400) 3 \times 10^8 \approx 2 \times 10^{12}$ meters. This is a ratio of horizontal arc length to vertical distance of about 1/20, which is fairly small. It doesn't look nearly as small as the ratio we derived above for the Earth, but we can't really compare those directly. To do a direct comparison, we would have to look at an object orbiting the Earth; for example, in low Earth orbit an object completes one full orbit in about 90 minutes or 5400 seconds or about $10^{11}$ meters, and travels an arc length of $2 \pi r \approx 4 \times 10^7$ meters. This gives a ratio of about 1/2500, which is still quite a bit smaller than that for the Sun even at the radius of the Earth's orbit.

If a test particle is released at the Earth's position withe zero tangential velocity it will travel toward the sun. Is this also a geodesic path

Yes.

does the geodesic path depend on the velocity of the test particle relative to the sun?

Yes.

 

[A.T.]

If I toss a ball five meters high and five meters horizontally

I would suggest to start with a purely vertical toss. With only one spatial dimension you have a 2D space-time with is simpler to visualize.

This is my own non-animated way of looking at it:

• A. Two inertial particles, at rest relative to each other, in flat spacetime (i.e. no gravity), shown with inertial coordinates. Drawn as a red distance-time graph on a flat piece of paper with blue gridlines.
  
• B₁. The same particles in the same flat spacetime, but shown with non-inertial coordinates. Drawn as the same distance-time graph on an identical flat piece of paper except it has different gridlines.
  
  B₂. Take the flat piece of paper depicted in B₁, cut out the grid with some scissors, and wrap it round a cone. Nothing within the intrinsic geometry of the paper has changed by doing this, so B₂ shows exactly the same thing as B₁, just presented in a different way, showing how the red lines could be perceived as looking "curved" against a "straight" grid.
  
• C. Two free-falling particles, initially at rest relative to each other, in curved spacetime (i.e. with gravity), shown with non-inertial coordinates. This cannot be drawn to scale on a flat piece of paper; you have to draw it on a curved surface instead. Note how C looks rather similar to B₂. This is the equivalence principle in action: if you zoomed in very close to B₂ and C, you wouldn't notice any difference between them.

Note the diagrams above aren't entirely accurate because they are drawn with a locally-Euclidean geometry, when really they ought to be drawn with a locally-Lorentzian geometry. I've drawn it this way as an analogy to help visualise the concepts.

Is there any physical significance to the ct increment of 6 trillion meters or is this just a mathematical byproduct to be ignored?

Maybe you could say that it indicates a rather small space-time distortion, since such a large increment along time, produces so little deviation towards the spatial direction. Note that the diagrams in the videos above are not to scale for an apple falling on Earth, to better show (exaggerate) the distortion.

Is this also a geodesic path and does the geodesic path depend on the velocity of the test particle relative to the sun?

Yes, the geodesic depends on the initial velocity of the particle, which defines it's initial "direction" in space-time.

 

[Bob R]

Thank you to PeterDonis and A.T. I wasn't expecting such rapid detailed answers. It will take me a bit longer to digest your remarks.

Thanks so much,

Bob R.

 

[Bob R]

I've been thinking about the worldline for the sample ball tossed 5 meters in the air over two seconds. I have to wonder about the ct component along the time axis being 6 e8 meters and what this means physically. It suggests that what we witness of this event is only the spatial movement of the ball, and that movement is very small compared to the time axis displacement. Here goes a wild silly speculation:

If one were to consider what we witness of the universe from the big bang 13.7 billion years ago as being analogous to the ball's spatial movement over two seconds, then there is a corresponding worldline movement along the time axis of (3 e8 m/sec) x (13.7 e9 years x 365.24 days/year x 24 hours/day x 3600 sec/hour) ~ 1.296981 e26 m. Is it possible that we humans, limited by witnessing events unfolding at no more than the speed of light, are part of a cosmos vastly larger than our universe?

 

[Nugatory]

I've been thinking about the worldline for the sample ball tossed 5 meters in the air over two seconds. I have to wonder about the ct component along the time axis being 6 e8 meters and what this means physically. It suggests that what we witness of this event is only the spatial movement of the ball, and that movement is very small compared to the time axis displacement.

Do remember that we're also moving along the time axis (time is passing for us too) as we're watching the ball. If I throw a ball five meters into the air and then catch it when it comes down two seconds later:
- The ball leaves my hand at one point in spacetime; we'll call that point A.
- I move $6\times{10}8$ meters through spacetime to point B, where the ball returns to my hand. That path is not a geodesic because the surface of the Earth is pushing me.
- The ball is moving on a different path between points A and B, one which is a geodesic. That path is very close to mine, but not exactly the same; at its greatest distance it is 5 meters away from my path.

It's true that 5 meters isn't much compared with $6\times{10}8$ meters, but all that means is that the Earth's gravity is quite weak so the ball was moving very slowly compared with the speed of light. Perform the same experiment on the surface of a neutron star, and you'll find that distance we move between points A and B is quite a bit less than the maximum separation between our worldline and the worldline of the ball.

 

[Bob R]

Thank you for your answer, Nugatory. I understand from what you say that the ball's wordline and the 4-D path of the person tossing it are nearly the same. They deviate from one another by at most 5 m. What we witness of the ball's worldline is the 3-D projection of that line in space.

I can't help but wonder about the significance of what we don't see, the 6 e8 m along the ct axis. As you point out, it shows that we move much slower than the speed of light. If what we witness in our world were near the speed of light, the worldline would have spatial and ct components much closer together. However, we humans would not be able to track events at such speed. Are we missing events along the ct axis because we cannot see them?

Best wishes,

Bob R

 

[Dale]

Is it possible that we humans, limited by witnessing events unfolding at no more than the speed of light, are part of a cosmos vastly larger than our universe?

The idea you have is likely to be correct. The way that we would say it is that the size of the observable universe is probably rather small compared to the size of the universe.

 

[Bob R]

Thank you, Dale.

Amazing!

Bob R